A mirror-edge figure is a figure whose every edge is bisected perpendicularly by a mirror. These form the bulk of the uniform figures. This is a short note to show how I find the circumdiameter of any given mirror-edge figure, given its symbol in my notation. It involving creating a matrix, and multiplying it by a vector, and then taking the dot product of the two vectors.
Identify the edges and lengths. For this exercise, we use the truncated 221, where the remaining edges are half the edge of the new edges old vertex figure.
* This is how the new edges sit in relation
2 to the old triangles of the 2_21.
------
1 / \
\ /
\ /
* ---- *
The "icosahedral" form of 221 is "{3,3,3,3,B}" or SSSSB. The edges are /S/SSSB, or (1 S 2 S 0 S 0 S 0 B 0) Here the edge lengths are shown to the nodes, and the old form (which identifies branches as letters) serve as commas.
The matrix is constructed for SSSSB. All matrix constructions follow the same manner: a series and an "animal" are put into the form, and then the rest is populated by arithmetic infil.
The series are according to the end sequences.
...A 1 1 2 2 2 2 2 2 .... ...B 3 2 4 6 5 4 3 2 1 0 ...S 1 2 3 4 5 6 7 8 9 10 ...Q r2 2 2 2 2 2 2 2 .... r2 = 1.414... ..QS r2 2r2 3 2 1 0 r2 = 1.414... ...F f 2 3-f 4-2f 5-3f f = 1.618... ...H r3 2 1 0 r3 = 1.732...
The animals are...
For For For For
S, Q, F, H A QS B
[ n ] [ n n-2 ] [ 2n-2 n-1 ] [2n-2 n-1 2n-6]
[n-2 n ] [ n-1 4 ] [ n-1 4 n-3]
[2n-6 n-3 n ]
The "animal" is written in the bottom right hand corner of the matrix. The series starts from the bottom left and works its way to the top. The next element of the series is the denominator. We're doing a six dimensional figure here, so:
+-- 3
| [ 4 . . . . . ] ( 1 )
| [ 5 : . . . . ] ( 2 )
2 | [ 6 . : . . . ] ( 0 )
- | [ 4 . . 10 5 6 ] ( 0 )
3 <-+ [ 2 . . 5 4 3 ] ( 0 )
[ 3 . . 6 3 6 ] ( 0 )
The numerator is Always "2" . You can see here, the animal B evaluated for n=6. Column A is the series, continued past the end of the matrix, the top number becoming the denominator of the matrix.
The Column vector (1,2,0,0,0,0) represents the figure we propose to find the diameter of.
The next step is to fill in the dots. The matrix is symmetric (ie Aij=Aji). And then Aij= j.Ai1. So for a given row, that begins "x", we write x 2x 3x 4x until the diagonal or the animal is met.
[ 4 . . . . . ] ( 1 )
[ 5 10 . . . . ] ( 2 )
2 [ 6 12 18 . . . ] ( 0 )
- [ 4 8 12 10 5 6 ] ( 0 )
3 [ 2 4 6 5 4 3 ] ( 0 )
[ 3 6 9 6 3 6 ] ( 0 )
And because it is symmetric, the top is filled out as well. Having got such a matrix and vector, we multiply these to yield a second vector (Y), and then take the dot product of the two vectors (XY).
[ 4 5 6 4 2 3 ] ( 1 ) (14) 14 = 1*14
[ 5 10 12 8 4 6 ] ( 2 ) (25) 50 = 2*25
2 [ 6 12 18 12 6 9 ] ( 0 ) (30) 0 = 0*30
- [ 4 8 12 10 5 6 ] ( 0 ) (20) 0 = 0*30
3 [ 2 4 6 5 4 3 ] ( 0 ) (10) 0 = 0*30
[ 3 6 9 6 3 6 ] ( 0 ) (15) 0 = 0*30
total: 64
The circumdiameter of this figure is then sqrt(64 * 2/3) where 2/3 is the fraction out the left, or 8/3 sqrt(6).
This works because one can define (using the fundemental reflective region), a co-ordinate system, so that the unit vector makes that figure have unit edge.
The vector represents a position-vector S = sum(j)(xj * Vj). The matrix represents the dot products of (Vi . Vj). This allows us to calculate the norm of S by multiplying the matrix Mij * Vi * Vj.
The construction remains a mystery. The denominator of the fraction (here the "3") is a very important function of the symmetry group, and is also the determinate of the matrix.
Even the construction of the series is something that I understand the process of, but can not explain elsewise. The diameter of any regular figure, or rectified figure, is simply 2(V1)(V2)(V3).../TF where V1, V2, V3 are the elements of a (prismatic) vertex figure, and TF is the total figure.
Why the matricies should yield a consistant construction is beyond me.
Copyright 2002 Wendy Krieger